# How do you solve the following system: 3x+2y =0, y + 4x = 16 ?

Jun 22, 2018

$x = \frac{32}{5} , y = - \frac{48}{5}$

#### Explanation:

By the first equation we get

$y = - \frac{3}{2} x$
plugging this in the second one:

$- \frac{3}{2} x + 4 x = 16$
$x = \frac{32}{5}$

$y = - \frac{48}{5}$

Jun 22, 2018

$x = 6 \left(\frac{2}{5}\right) , y = - 9 \left(\frac{3}{5}\right)$

#### Explanation:

$3 x + 2 y = 0$ (1)

$2 y = - 3 x$

$y + 4 x = 16$ (2)

Substituting y in terms of x in (2),

$- \left(\frac{3}{2}\right) x + 4 x = 16$

$- 3 x + 8 x = 32$

$5 x = 32$ or $x = 6 \left(\frac{2}{5}\right)$

$y = 16 - 4 x = 16 - \left(\frac{128}{5}\right) = - \frac{48}{5} = - 9 \left(\frac{3}{5}\right)$